BÀi 5
a) x^2 - 12 x + 11 = 0
b) 4x^2 - 4x - 3 = 0
c) 4x^2 - 12x - 7 = 0
d) x^3 - 6x^2 = 8 - 12x
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5 tháng 8 2019
a) \(x^2-12x+11=0\)
\(\Leftrightarrow x^2-2.6.x+36-25=0\)
\(\Leftrightarrow\left(x-6\right)^2-25=0\)
\(\Leftrightarrow\left(x-6\right)^2=25=5^2=\left(-5\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=5\\x-6=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=11\\x=1\end{matrix}\right.\)
Vậy : \(x\in\left\{11,1\right\}\)
c) \(4x^2-12x-7=0\)
\(\Leftrightarrow\left(2x\right)^2-2.2x.3+9-16=0\)
\(\Leftrightarrow\left(2x-3\right)^2-16=0\)
\(\Leftrightarrow\left(2x-3\right)^2=16=4^2=\left(-4\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=7\\2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)
Vậy : \(x\in\left\{\frac{7}{2},-\frac{1}{2}\right\}\)
Câu b) và d) xíu em làm sau, em hơi bận chút !!
5 tháng 8 2019
Làm tiếp nha >>>
b) \(4x^2-4x-3=0\)
\(\Leftrightarrow\left(2x\right)^2-2.2x.1+1-4=0\)
\(\Leftrightarrow\left(2x-1\right)^2-4=0\)
\(\Leftrightarrow\left(2x-1\right)^2=4=2^2=\left(-2\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)
Vậy : \(x\in\left\{\frac{3}{2},-\frac{1}{2}\right\}\)
d) \(x^3-6x^2=8-12x\)
\(\Leftrightarrow x^3-6x^2-\left(8-12x\right)=0\)
\(\Leftrightarrow x^3-6x^2-8+12x=0\)
\(\Leftrightarrow x^3-3.x^2.2+3.x.2^2-2^3=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy : \(x=2\)
P/s : Hằng đẳng thức với lập phương khó thật, rối câu d) mãi mới nghĩ ra >>
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13 tháng 12 2021
c: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
11 tháng 10 2020
a) \(x\left(x-2\right)-7x+14=0\)
\(\Leftrightarrow x\left(x-2\right)-7\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=7\end{cases}}\)
b) \(x^2\left(x-3\right)+12-4x=0\)
\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x^2=4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\pm2\end{cases}}\)
c) \(x^2+12x-13=0\)
\(\Leftrightarrow\left(x^2-x\right)+\left(13x-13\right)=0\)
\(\Leftrightarrow x\left(x-1\right)+13\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+13\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-13\end{cases}}\)
d) \(4x^2-4x=8\)
\(\Leftrightarrow x^2-x-2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
e) \(x^2-6x=1\)
\(\Leftrightarrow\left(x-3\right)^2=10\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=\sqrt{10}\\x-3=-\sqrt{10}\end{cases}}\Rightarrow\orbr{\begin{cases}x=3+\sqrt{10}\\x=3-\sqrt{10}\end{cases}}\)
11 tháng 10 2020
a) x( x - 2 ) - 7x + 14 = 0
<=> x( x - 2 ) - 7( x - 2 ) = 0
<=> ( x - 2 )( x - 7 ) = 0
<=> \(\orbr{\begin{cases}x-2=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=7\end{cases}}\)
b) x2( x - 3 ) + 12 - 4x = 0
<=> x2( x - 3 ) - 4( x - 3 ) = 0
<=> ( x - 3 )( x2 - 4 ) = 0
<=> \(\orbr{\begin{cases}x-3=0\\x^2-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\pm2\end{cases}}\)
c) x2 + 12x - 13 = 0
<=> x2 - x + 13x - 13 = 0
<=> x( x - 1 ) + 13( x - 1 ) = 0
<=> ( x - 1 )( x + 13 ) = 0
<=> \(\orbr{\begin{cases}x-1=0\\x+13=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-13\end{cases}}\)
d) 4x2 - 4x = 8
<=> 4( x2 - x ) = 8
<=> x2 - x = 2
<=> x2 - x - 2 = 0
<=> x2 + x - 2x - 2 = 0
<=> x( x + 1 ) - 2( x + 1 ) = 0
<=> ( x + 1 )( x - 2 ) = 0
<=> \(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
e) x2 - 6x = 1
<=> x2 - 6x + 9 = 1 + 9
<=> ( x - 3 )2 = 10
<=> ( x - 3 )2 = ( ±√10 )2
<=> \(\orbr{\begin{cases}x-3=\sqrt{10}\\x-3=-\sqrt{10}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3+\sqrt{10}\\x=3-\sqrt{10}\end{cases}}\)
27 tháng 8 2020
Bài 1.
a) ( 7x - 3 )2 - 5x( 9x + 2 ) - 4x2 = 18
<=> 49x2 - 42x + 9 - 45x2 - 10x - 4x2 = 18
<=> -52x + 9 = 18
<=> -52x = 9
<=> x = -9/52
b) ( x - 7 )2 - 9( x + 4 )2 = 0
<=> x2 - 14x + 49 - 9( x2 + 8x + 16 ) = 0
<=> x2 - 14x + 49 - 9x2 - 72x - 144 = 0
<=> -8x2 - 86x - 95 = 0
<=> -8x2 - 10x - 76x - 95 = 0
<=> -8x( x + 5/4 ) - 76( x + 5/4 ) = 0
<=> ( x + 5/4 )( -8x - 76 ) = 0
<=> \(\orbr{\begin{cases}x+\frac{5}{4}=0\\-8x-76=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{4}\\x=-\frac{19}{2}\end{cases}}\)
c) ( 2x + 1 )2 + ( 4x - 1 )( x + 5 ) = 36
<=> 4x2 + 4x + 1 + 4x2 + 19x - 5 = 36
<=> 8x2 + 23x - 4 - 36 = 0
<=> 8x2 + 23x - 40 = 0
=> Vô nghiệm ( lớp 8 chưa học nghiệm vô tỉ nghen ) :))
Bài 2.
a) x2 - 12x + 39 = ( x2 - 12x + 36 ) + 3 = ( x - 6 )2 + 3 ≥ 3 > 0 ∀ x ( đpcm )
b) 17 - 8x + x2 = ( x2 - 8x + 16 ) + 1 = ( x - 4 )2 + 1 ≥ 1 > 0 ∀ x ( đpcm )
c) -x2 + 6x - 11 = -( x2 - 6x + 9 ) - 2 = -( x - 3 )2 - 2 ≤ -2 < 0 ∀ x ( đpcm )
d) -x2 + 18x - 83 = -( x2 - 18x + 81 ) - 2 = -( x - 9 )2 - 2 ≤ -2 < 0 ∀ x ( đpcm )
NL
1
12 tháng 12 2023
a: \(x^2+12x+36=0\)
=>\(x^2+2\cdot x\cdot6+6^2=0\)
=>\(\left(x+6\right)^2=0\)
=>x+6=0
=>x=-6
b: \(4x^2-4x+1=0\)
=>\(\left(2x\right)^2-2\cdot2x\cdot1+1^2=0\)
=>\(\left(2x-1\right)^2=0\)
=>2x-1=0
=>2x=1
=>x=1/2
c: \(x^3+6x^2+12x+8=0\)
=>\(x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=0\)
=>\(\left(x+2\right)^3=0\)
=>x+2=0
=>x=-2
NT
1
NT
1
TA
1 tháng 8 2017
a) \(5\left(x+3\right)-6x-2x^2=0\) \(\Leftrightarrow5.\left(x+3\right)-2x.\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(5-2x\right)=0\Leftrightarrow\hept{\begin{cases}x+3=0\\5-2x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-3\\2x=5\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-3\\x=\frac{5}{2}\end{cases}}}\)
b) \(6x.\left(x^2-2\right)-\left(2-x^2\right)=0\) \(\Leftrightarrow6x.\left(x^2-2\right)+\left(x^2-2\right)=0\)
\(\Leftrightarrow\left(x^2-2\right)\left(6x+1\right)=0\Leftrightarrow\hept{\begin{cases}x^2-2=0\\6x+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2=2\\6x=-1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\sqrt{2}\\x=\frac{-1}{6}\end{cases}}}\)
c) \(4x.\left(x-2017\right)-x+2017=0\) \(\Leftrightarrow4x.\left(x-2017\right)-\left(x-2017\right)=0\)
\(\Leftrightarrow\left(x-2017\right).\left(4x-1\right)=0\) \(\Leftrightarrow\hept{\begin{cases}x-2017=0\\4x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2017\\4x=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2017\\x=\frac{1}{4}\end{cases}}}\)
d) \(12x=x^2+36\) \(\Leftrightarrow x^2-12x+36=0\) \(\Leftrightarrow\left(x-6\right)^2=0\) \(\Rightarrow x-6=0\) \(\Leftrightarrow x=6\)
a) \(x^2-12x+11\)\(=0\)
\(\Leftrightarrow\left(x-6\right)^2-25=0\)
\(\Leftrightarrow\left(x-6+5\right)\left(x-6-5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=11\end{matrix}\right.\)
a)\(x^2-12x+11=0\)
\(x^2-x-11x+11=0\)
\(\left(x^2-x\right)-\left(11x-11\right)=0\)
\(x\left(x-1\right)-11\left(x-1\right)=0\)
\(\left(x-1\right)\left(x-11\right)=0\)
\(=>\left[{}\begin{matrix}x-1=0\\x-11=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=1\\x=11\end{matrix}\right.\)
b)\(4x^2-4x-3=0\)
\(4x^2-2x+6x-3=0\)
\(2x\left(2x-1\right)+3\left(3x-1\right)=0\)
\(\left(2x-1\right)\left(2x+3\right)=0\)
\(=>\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=0,5\\x=-1,5\end{matrix}\right.\)\
c)\(4x^2-12x-7=0\)
\(4x^2-14x+2x-7=0\)
\(2x\left(2x-7\right)+\left(2x-7\right)=0\)
\(\left(2x-7\right)\left(2x+1\right)=0\)
\(=>\left[{}\begin{matrix}2x-7=0\\2x+1=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.\)